I want to understand this and need a little help. As I play a few with spice
it seems to give a voltage about a few ten mV but not really like a
reference source. What is the benefit of this schematic and how does it
work?

I want to understand this and need a little help. As I play a few with
spice it seems to give a voltage about a few ten mV but not really like a
reference source. What is the benefit of this schematic and how does it
work?

Marte

Click to expand...

Hello Marte,

This circuit may be as bad as it looks nowadays.

Prerequisite:
A Si-transistor has a Vbe of about 0.7V with a tempco of
about -1.8mV/degree.
A Ge-transistor has a Vbe of about 0.36V* with a tempco of -1.xmV/degree**.

The concept of this circuit:

Vbe of Q2(2N2222) is at about 0.7V with a tempco of -1.8mV/degree
regardless of Q1.
The base of Q1 is 0.36V* below this 0.7V at 0.34V (0.7-0.36).
The net effect is that the voltage Ve_Q1 at the emitter of Q1 is 0.34V with
a tempco of 0mV/degree if the Vbe_tempco of both transistors is the same.

Ve_Q1=Vbe_Q2-Vbe_Q1=0.7-0.36=0.34V

Ve_tempco=Vbe_tempcoQ2-Vbe_tempco_Q1=-1.8mV-(-1.xmV)=-0.ymV.

The opamp circuit around requires a potentiometer to adjust the output
voltage because of the wide variation of Vbe.

My advice: Don't use this circuit for a commercial application because
Ge-transistors are dead since 30 years.
I also have never seen Ge-transistors in SMT.

Best regards,
Helmut

* Just a guess. The real value may have a wide variation in the datasheets.
** -1.xmV means -1.5mV or something like that. Ideally it should be equal
the tempco of the Si-transistor.

Prerequisite:
A Si-transistor has a Vbe of about 0.7V with a tempco of
about -1.8mV/degree.
A Ge-transistor has a Vbe of about 0.36V* with a tempco of -1.xmV/degree**.

The concept of this circuit:

Vbe of Q2(2N2222) is at about 0.7V with a tempco of -1.8mV/degree
regardless of Q1.
The base of Q1 is 0.36V* below this 0.7V at 0.34V (0.7-0.36).
The net effect is that the voltage Ve_Q1 at the emitter of Q1 is 0.34V with
a tempco of 0mV/degree if the Vbe_tempco of both transistors is the same.

Ve_Q1=Vbe_Q2-Vbe_Q1=0.7-0.36=0.34V

Ve_tempco=Vbe_tempcoQ2-Vbe_tempco_Q1=-1.8mV-(-1.xmV)=-0.ymV.

The opamp circuit around requires a potentiometer to adjust the output
voltage because of the wide variation of Vbe.

My advice: Don't use this circuit for a commercial application because
Ge-transistors are dead since 30 years.
I also have never seen Ge-transistors in SMT.

Best regards,
Helmut

* Just a guess. The real value may have a wide variation in the datasheets.
** -1.xmV means -1.5mV or something like that. Ideally it should be equal
the tempco of the Si-transistor.

Click to expand...

That's getting close but not quite there. Using your notation for Q1 and
Q2, you can see that Vout is Vce,sat of Q2 and nicely at low impedance.
Q2 must saturate and Q1 is diode connected with its Vce=Vbe,Q2 for
practical purposes. The Vce,sat is inherently the difference between two
forward biased PN junctions so that on that basis alone you would expect
the tempco to be an order of magnitude below that of single diode. Also,
when Q2 saturates, the BE junctions of Q1 and Q2 are in parallel, with
Q1 providing negative feedback by shunting the supply current from
driving the base of Q2 into its CE circuit. The net effect is to produce
a Vce,sat with regulated base drive. It is not necessary for Q1 to be
Germanium, an Si will work too but the Vout is of lower magnitude.

That's getting close but not quite there. Using your notation for Q1 and
Q2, you can see that Vout is Vce,sat of Q2 and nicely at low impedance. Q2
must saturate and Q1 is diode connected with its Vce=Vbe,Q2 for practical
purposes. The Vce,sat is inherently the difference between two forward
biased PN junctions so that on that basis alone you would expect the
tempco to be an order of magnitude below that of single diode. Also, when
Q2 saturates, the BE junctions of Q1 and Q2 are in parallel, with Q1
providing negative feedback by shunting the supply current from driving
the base of Q2 into its CE circuit. The net effect is to produce a Vce,sat
with regulated base drive. It is not necessary for Q1 to be Germanium, an
Si will work too but the Vout is of lower magnitude.

Click to expand...

Hello Fred,

Sorry, your assumption is wrong. The basic idea is to compensate the tempco
of Vbe1 wth the tempco of Vbe2 by using Q2 in its active region with Vce >
Vce_sat.
This can only be achieved with a Germanium transistor for Q1 with a much
lower
Vbe than the Vbe of Q2.

Sorry, your assumption is wrong. The basic idea is to compensate the tempco
of Vbe1 wth the tempco of Vbe2 by using Q2 in its active region with Vce >
Vce_sat.
This can only be achieved with a Germanium transistor for Q1 with a much
lower
Vbe than the Vbe of Q2.

Best regards,
Helmut

Click to expand...

Why does the National data sheet show Q1 as being either Si or Ge then?

Right, all clear. With Ge transistors it is clear. With Si transistors the
Voltage will be very low and sensible to Vcc.

May be they wanted to tell, that LM311 may also be useable as a opamp and
lead to its ability of common mode input below gnd.

What's about Schotthy Diode instead of Germanium transistor? Or LED
instead of 2N2222?

Marte

Click to expand...

Hello Marte,

Both methods may be at least as good as the combination of a Ge-transistor
and a Si-transistor.
The ladder may be better because it gives a higher voltage. You have to
"play" with LEDs
of different color and from different manufacturers of course.
Just try it in an oven.

Q2, you can see that Vout is Vce,sat of Q2 and nicely at low impedance. Q2
must saturate

Click to expand...

right here. But guess Vce,sat with a few mV in case of two Si-transistors?

The Vce,sat is inherently the difference between two forward biased PN
junctions so that on that basis alone you would expect the tempco to be an
order of magnitude below that of single diode.

Click to expand...

But I get the tempco with about .x mV/K at a voltage of a few ten mV instead
of 500 to 600 mV from a simple si-diode. I would say that wouldn't be a
real benefit then.

when Q2 saturates, the BE junctions of Q1 and Q2 are in parallel, with Q1
providing negative feedback by shunting the supply current from driving
the base of Q2 into its CE circuit.

Click to expand...

At least here the simulations with LTSPICE disagrees with you. Making a DC
sweep with V1 creates a nearly linear function of Ib(Q2)

The net effect is to produce a Vce,sat with regulated base drive. It is
not necessary for Q1 to be Germanium, an Si will work too but the Vout is
of lower magnitude.

Click to expand...

In my simulation it looks like that changing the more powerfol transistor to
the upper side (Q1) and the "smaller" one (I took 2N3904 for simulations
now) as Q2 produce more stable results. I think this stabilases Helmuts
theorie, isn't ist?

Keep in mind that d(Vbe)/dT is not constant; you can adjust it by
changing i(c), possibly achieving better cancellation / temperature
compensation. Also, Vce(sat) has a small positive tempco, which can
be handy too.

And the circuit does indeed work with both Qs as Si but with (obviously)
a much lower voltage as FB details.

Click to expand...

We agree the circuit works per Fred's description when Q1 is
silicon, and we agree the output voltages are quite different for
silicon versus germanium Q1. I measure Uo = 18.18mV with Q2=PN2222a,
Q1=KTC3198, with drift of roughly 0.3% / C, line regulation = 0.9% /
V. With Q1=2n5772, Uo = 6.88mV.

However, the application circuits then make no sense, as this
massive a change in reference voltage would obviously have a huge
impact on the "Low Voltage Adjustable Reference Supply" and the
precision waveform squarer I identified earlier. The reference
voltage would then be only a few times the comparator's offset
voltage, for one thing, and massively dependent on Vbe matching, for
another.

Since the LM311 was originally a National Semiconductor design
("LMxxx," after all) I conclude T.I. copied National's datasheet, that
some helpful T.I. app engineer noticed the obsolete transistor, and
erred in recommending a substitute Q1. Perhaps he's the same fellow
who left the "dot" off the precision squarer circuit?, which, as
drawn, produces no useful output!

We agree the circuit works per Fred's description when Q1 is
silicon, and we agree the output voltages are quite different for
silicon versus germanium Q1. I measure Uo = 18.18mV with Q2=PN2222a,
Q1=KTC3198, with drift of roughly 0.3% / C, line regulation = 0.9% /
V. With Q1=2n5772, Uo = 6.88mV.

However, the application circuits then make no sense, as this
massive a change in reference voltage would obviously have a huge
impact on the "Low Voltage Adjustable Reference Supply" and the
precision waveform squarer I identified earlier. The reference
voltage would then be only a few times the comparator's offset
voltage, for one thing, and massively dependent on Vbe matching, for
another.

Since the LM311 was originally a National Semiconductor design
("LMxxx," after all) I conclude T.I. copied National's datasheet, that
some helpful T.I. app engineer noticed the obsolete transistor, and
erred in recommending a substitute Q1. Perhaps he's the same fellow
who left the "dot" off the precision squarer circuit?, which, as
drawn, produces no useful output!

Click to expand...

You could well be right. I never said it was a good circuit, just ageed
about it *could* (indeed does) work with silicon Qs. Well spotted re
the "dot"!

No, at lest in this datasheet ST correctly shows the voltage
reference application circuit (pg. 9) with a 2n1304, which is
germanium:
T.I. is alone in showing a silicon transistor, which doesn't work in

Try Q1 the more powerful transistor and Q2 the smaller ones. So Uo will be a
little better.

However, the application circuits then make no sense,

Click to expand...

I agree, one look on the offset voltage and all the dreams are gone

precision waveform squarer I identified earlier.

Click to expand...

This may be the next topic: What is a "precision squarer" for?

Since the LM311 was originally a National Semiconductor design
("LMxxx," after all) I conclude T.I. copied National's datasheet, that
some helpful T.I. app engineer noticed the obsolete transistor, and
erred in recommending a substitute Q1. Perhaps he's the same fellow
who left the "dot" off the precision squarer circuit?, which, as
drawn, produces no useful output!

Click to expand...

Right, I saw so many application hints in datasheets, with obvious mistakes
;-)

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